College Chemistry Quiz: Introduction For Reactions

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Introduction For ReactionsQuestion 1 of 20

A student performs a combustion reaction by burning 2.50 g of methane ( $CH_4$ ) in excess oxygen. If the theoretical yield of carbon dioxide is 6.88 g, but the student actually recovers 5.85 g of $CO_2$ , what is the percent yield of this reaction?

37.2%

42.8%

85.0%

117.6%

235.2%

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College Chemistry Quiz

College Chemistry Quiz: Introduction For Reactions

Practice Introduction For Reactions in College Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Introduction For Reactions, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

37.2%
42.8%
85.0% (correct answer)
117.6%
235.2%

Explanation: When you encounter percent yield problems, you're working with the relationship between what should theoretically happen in a reaction versus what actually occurs in the lab. Percent yield equals actual yield divided by theoretical yield, multiplied by 100%. The problem gives you everything you need: the theoretical yield is 6.88 g of

CO_2

, and the actual yield recovered is 5.85 g. Applying the formula: Percent yield =

\frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% = \frac{5.85 \text{ g}}{6.88 \text{ g}} \times 100\% = 85.0\%

This confirms answer C is correct. Looking at the wrong answers: A) 37.2% would result from incorrectly dividing 2.50 g (the methane mass) by 6.88 g, but percent yield compares actual product to theoretical product, not reactant to product. B) 42.8% appears to come from dividing the methane mass by the actual

CO_2

recovered (2.50/5.85), which again uses the wrong values entirely. D) 117.6% would mean you recovered more product than theoretically possible, which would require flipping the fraction (6.88/5.85) – this represents a fundamental misunderstanding since percent yields above 100% indicate calculation errors in most cases. Remember: percent yield problems always follow the same pattern – find what you actually got, divide by what you should have gotten theoretically, then multiply by 100%. The theoretical yield is your benchmark for a "perfect" reaction.

Question 2

When balancing the redox equation $Cr_2O_7^{2-} + Fe^{2+} + H^+ \rightarrow Cr^{3+} + Fe^{3+} + H_2O$ in acidic solution, what is the coefficient of $H^+$ in the balanced equation?

6
8
12
14 (correct answer)
16

Explanation: When you encounter redox equations in acidic solution, you need to balance both mass and charge by systematically balancing each half-reaction separately, then combining them. Start with the reduction half-reaction:

Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O

. The dichromate ion gains 6 electrons (chromium goes from +6 to +3 oxidation state), requires 14

H^+

ions to balance the oxygen atoms, and produces 7 water molecules. The oxidation half-reaction is simpler:

Fe^{2+} \rightarrow Fe^{3+} + e^-

. Each iron ion loses one electron. To balance electrons, multiply the iron half-reaction by 6:

6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^-

Combining both half-reactions:

Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O

The coefficient of

H^+

is 14, making answer D correct. Choice A (6) likely comes from incorrectly using the number of electrons transferred. Choice B (8) might result from miscounting oxygen atoms or forgetting that dichromate contains two chromium atoms. Choice C (12) could arise from arithmetic errors when balancing the chromium atoms or water molecules. Remember the systematic approach: separate into half-reactions, balance atoms other than H and O first, then add

H_2O

for oxygen and

H^+

for hydrogen in acidic solutions. Always verify your final equation balances both atoms and charge.

Question 3

A solution contains 0.150 M $NaCl$ and 0.200 M $AgNO_3$ . When these solutions are mixed in equal volumes, which statement best describes the reaction outcome?

$NaCl$ is the limiting reactant, and 0.075 M $AgCl$ precipitates (correct answer)
$AgNO_3$ is the limiting reactant, and 0.100 M $AgCl$ precipitates
$NaCl$ is the limiting reactant, and 0.150 M $AgCl$ precipitates
Both reactants are consumed completely with no excess remaining
$AgNO_3$ is the limiting reactant, and 0.200 M $AgCl$ precipitates

Explanation: This question tests your understanding of limiting reactants and stoichiometry in precipitation reactions. When you see two ionic compounds that can form a precipitate, you need to determine which reactant will be completely consumed first. The balanced equation is:

NaCl + AgNO_3 \rightarrow AgCl(s) + NaNO_3

. Since the stoichiometry is 1:1, you need equal moles of each reactant. When equal volumes are mixed, each solution is diluted by half. The new concentrations become:

NaCl

: 0.075 M and

AgNO_3

: 0.100 M. Since you need equal moles and

NaCl

has the lower concentration, it's the limiting reactant. All 0.075 M of

NaCl

will react to form 0.075 M

AgCl

precipitate, with 0.025 M

AgNO_3

remaining unreacted. Option A correctly identifies

NaCl

as limiting and gives the right precipitate concentration. Option B incorrectly claims

AgNO_3

is limiting—it's actually in excess. Option C makes the error of using the original

NaCl

concentration (0.150 M) instead of the diluted concentration after mixing. Option D wrongly suggests both reactants are completely consumed, ignoring that

AgNO_3

is in excess. Remember: when solutions are mixed, always account for dilution first, then apply stoichiometry. The limiting reactant is the one that produces fewer moles of product, and in precipitation problems, this determines how much solid forms.

Question 4

Which of the following reactions represents a double displacement (metathesis) reaction?

$Zn(s) + CuSO_4(aq) \rightarrow ZnSO_4(aq) + Cu(s)$
$CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$
$AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)$ (correct answer)
$2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$
$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$

Explanation: When you encounter questions about reaction types, focus on identifying the key structural changes happening to the compounds involved. Double displacement (metathesis) reactions have a very specific pattern: two ionic compounds exchange their ions to form two new compounds, following the general form AB + CD → AD + CB. Option C demonstrates this perfectly:

AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)

. Here, the silver ion (Ag⁺) from silver nitrate combines with the chloride ion (Cl⁻) from sodium chloride, while the sodium ion (Na⁺) pairs with the nitrate ion (NO₃⁻). The ions have literally "switched partners" to form two new compounds. Option A represents a single displacement reaction, where zinc metal displaces copper from copper sulfate. Only one element is being replaced, not a complete ion exchange between two compounds. Option B shows a decomposition reaction, where one compound (calcium carbonate) breaks down into two simpler substances (calcium oxide and carbon dioxide). No ion exchange occurs here. Option D illustrates a synthesis or combination reaction, where two elements combine to form a single compound (water). This is the opposite of decomposition. To master reaction classification, memorize the double displacement pattern: look for two ionic compounds as reactants that form two new ionic compounds as products, with the positive and negative ions switching places. The telltale sign is often the formation of a precipitate, gas, or water as one of the products.

Question 5

In the oxidation-reduction reaction $MnO_4^- + Fe^{2+} \rightarrow Mn^{2+} + Fe^{3+}$ in acidic solution, which species undergoes reduction and what is the change in oxidation state?

$Fe^{2+}$ undergoes reduction with a change of -1 in oxidation state
$MnO_4^-$ undergoes reduction with a change of -5 in oxidation state (correct answer)
$Fe^{2+}$ undergoes reduction with a change of +1 in oxidation state
$MnO_4^-$ undergoes reduction with a change of -1 in oxidation state
Both species undergo reduction with equal changes in oxidation state

Explanation: When you encounter redox reactions, you need to track oxidation states to determine which species is oxidized (loses electrons) and which is reduced (gains electrons). The key is identifying how the oxidation states change from reactants to products. Let's analyze each species in this reaction. For

MnO_4^-

, manganese has an oxidation state of +7 (since oxygen is -2, and the overall charge is -1: x + 4(-2) = -1, so x = +7). When it becomes

Mn^{2+}

, the oxidation state drops from +7 to +2, a change of -5. This decrease in oxidation state means

MnO_4^-

gains electrons and undergoes reduction. For iron,

Fe^{2+}

has an oxidation state of +2, which increases to +3 in

Fe^{3+}

. This +1 change means iron loses an electron and is oxidized. Choice B correctly identifies that

MnO_4^-

undergoes reduction with a -5 change in oxidation state. Choice A incorrectly states that

Fe^{2+}

undergoes reduction when it actually undergoes oxidation, and the change is +1, not -1. Choice C correctly identifies the +1 change for iron but wrongly calls it reduction instead of oxidation. Choice D correctly identifies

MnO_4^-

as undergoing reduction but gives the wrong magnitude of change (-1 instead of -5). Remember: reduction means gaining electrons and decreasing oxidation state, while oxidation means losing electrons and increasing oxidation state. Always calculate the actual numerical change in oxidation states to verify your answer.

Question 6

When aqueous solutions of $Ba(NO_3)_2$ and $Na_2SO_4$ are mixed, a white precipitate forms. What is the net ionic equation for this reaction?

$Ba^{2+}(aq) + NO_3^-(aq) + Na^+(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s) + NaNO_3(aq)$
$Ba(NO_3)_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaNO_3(aq)$
$Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)$ (correct answer)
$Ba^{2+}(aq) + 2NO_3^-(aq) + 2Na^+(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s) + 2Na^+(aq) + 2NO_3^-(aq)$
$Ba(NO_3)_2(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s) + 2NO_3^-(aq)$

Explanation: When you encounter precipitation reactions, you need to identify what actually forms the precipitate and write the net ionic equation showing only the species that participate in forming the solid. First, let's understand what happens when

Ba(NO_3)_2

and

Na_2SO_4

mix. Both compounds dissociate completely in water, producing

Ba^{2+}

NO_3^-

Na^+

, and

SO_4^{2-}

ions. The white precipitate that forms is

BaSO_4

, which is insoluble according to solubility rules (most sulfates are soluble except those of barium, lead, and mercury). The net ionic equation includes only the ions that directly participate in forming the precipitate. Since

Ba^{2+}

and

SO_4^{2-}

combine to form solid

BaSO_4

, the net ionic equation is:

Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)

. This is answer C. Answer A is incorrect because it doesn't balance properly and mixes ionic and molecular formulas inappropriately. Answer B shows the complete molecular equation, not the net ionic equation—it includes all compounds rather than just the participating ions. Answer D represents the complete ionic equation, showing all dissociated ions including the spectator ions (

Na^+

and

NO_3^-

) that don't participate in the precipitation. Remember: net ionic equations strip away spectator ions and show only the chemical change occurring. Always identify the precipitate first using solubility rules, then write the equation using only the ions that form that precipitate.

Question 7

In the reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$ , if 5.0 mol of $N_2$ and 12.0 mol of $H_2$ are allowed to react, how many moles of $NH_3$ can be produced?

8.0 mol (correct answer)
10.0 mol
12.0 mol
15.0 mol
17.0 mol

Explanation: This is a limiting reagent problem, which requires you to determine which reactant will be completely consumed first and therefore limit the amount of product that can be formed. Start by examining the balanced equation:

N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)

. The stoichiometric ratios tell you that 1 mole of

N_2

reacts with 3 moles of

H_2

to produce 2 moles of

NH_3

. With 5.0 mol of

N_2

available, you would need

5.0 \text{ mol } N_2 \times \frac{3 \text{ mol } H_2}{1 \text{ mol } N_2} = 15.0 \text{ mol } H_2

. However, you only have 12.0 mol of

H_2

, so hydrogen is the limiting reagent. Using the limiting reagent (

H_2

) to calculate product:

12.0 \text{ mol } H_2 \times \frac{2 \text{ mol } NH_3}{3 \text{ mol } H_2} = 8.0 \text{ mol } NH_3

Choice A (8.0 mol) is correct. Choice B (10.0 mol) likely comes from incorrectly assuming

N_2

is limiting:

5.0 \times 2 = 10.0

. Choice C (12.0 mol) represents a common error of thinking all the

H_2

converts directly to

NH_3

without considering stoichiometry. Choice D (15.0 mol) might result from incorrectly multiplying the excess

N_2

by 3. Study tip: Always identify the limiting reagent first by calculating how much of one reactant you'd need based on the amount of the other. The reactant that "runs out first" determines your maximum product yield.

Question 8

When copper metal reacts with concentrated nitric acid, the products include copper(II) nitrate, nitrogen dioxide, and water. What is the coefficient of $HNO_3$ in the balanced equation?

2
4 (correct answer)
6
8
10

Explanation: When you encounter redox reactions involving metals and acids, you need to balance both mass and charge by carefully tracking electron transfer and ensuring all atoms are accounted for. Let's work through this systematically. Copper metal (

Cu

) reacts with concentrated nitric acid (

HNO_3

) to produce copper(II) nitrate (

Cu(NO_3)_2

), nitrogen dioxide (

NO_2

), and water (

H_2O

). The unbalanced equation is:

Cu + HNO_3 \rightarrow Cu(NO_3)_2 + NO_2 + H_2O

Start by balancing copper: 1

Cu

on each side. Next, notice that copper(II) nitrate contains 2 nitrate ions, so you need at least 2

HNO_3

molecules just for that. However, some nitric acid is also reduced to form

NO_2

. The balanced equation is:

Cu + 4HNO_3 \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O

Let's verify: 1 Cu, 4 N, 12 O, and 4 H on both sides. The charge is also balanced. Choice A (2) only accounts for the nitrate ions in the salt, ignoring the

NO_2

formation. Choice C (6) overcounts the nitrogen species needed. Choice D (8) severely overcounts, likely from doubling the actual requirement unnecessarily. The correct answer is B (4). Study tip: In metal-acid reactions, remember that some acid molecules provide ions for salt formation while others undergo reduction. Always count both roles when determining coefficients, and verify your balanced equation by checking both mass and charge balance.

Question 9

In the reaction $CaC_2(s) + 2H_2O(l) \rightarrow Ca(OH)_2(aq) + C_2H_2(g)$ , what volume of acetylene gas ( $C_2H_2$ ) at STP is produced from 16.0 g of calcium carbide ( $CaC_2$ )?

2.80 L
5.59 L (correct answer)
11.2 L
22.4 L
44.8 L

Explanation: This problem tests your ability to perform stoichiometric calculations involving gas volumes at standard temperature and pressure (STP). When you see a balanced chemical equation with specific masses and need to find gas volumes, think moles and molar ratios. Start by finding the molar mass of

CaC_2

40.08 + 2(12.01) = 64.10 \text{ g/mol}

. Convert the given mass to moles:

\frac{16.0 \text{ g}}{64.10 \text{ g/mol}} = 0.2496 \text{ mol } CaC_2

. From the balanced equation, the molar ratio shows that 1 mole of

CaC_2

produces 1 mole of

C_2H_2

. Therefore, 0.2496 mol of

CaC_2

produces 0.2496 mol of

C_2H_2

. At STP, one mole of any gas occupies 22.4 L. So:

0.2496 \text{ mol} \times 22.4 \text{ L/mol} = 5.59 \text{ L}

. This confirms answer B is correct. Answer A (2.80 L) represents exactly half the correct volume, suggesting an error in the molar ratio or calculation. Answer C (11.2 L) is exactly half of 22.4 L, indicating someone might have incorrectly assumed 0.5 moles of product. Answer D (22.4 L) would be correct if you had exactly 1 mole of

CaC_2

, but you're working with approximately 0.25 moles. Remember: always convert mass to moles first, apply the molar ratio from the balanced equation, then use the molar volume at STP (22.4 L/mol) for gas volume calculations.

Question 10

Which of the following best describes what happens when aqueous solutions of $FeCl_3$ and $NaOH$ are mixed?

A redox reaction occurs producing iron metal and chlorine gas
An acid-base neutralization occurs producing water and a salt solution
A precipitation reaction occurs forming insoluble iron(III) hydroxide (correct answer)
A single displacement reaction occurs with sodium replacing iron
No reaction occurs because both compounds are already ionic

Explanation: When you encounter a question about mixing ionic solutions, you need to predict what type of reaction will occur based on the chemical properties of the reactants and products. Let's analyze what happens when

FeCl_3

and

NaOH

mix. Iron(III) chloride dissociates into

Fe^{3+}

and

Cl^-

ions, while sodium hydroxide produces

Na^+

and

OH^-

ions. The key insight is recognizing that

Fe^{3+}

and

OH^-

can combine to form

Fe(OH)_3

, which is insoluble in water. The reaction is:

FeCl_3 + 3NaOH \rightarrow Fe(OH)_3(s) + 3NaCl

. This forms a precipitate, making answer C correct. Now let's examine why the other options are wrong. Option A suggests a redox reaction producing iron metal and chlorine gas, but neither

Fe^{3+}

nor

Cl^-

undergoes oxidation or reduction under these mild aqueous conditions. Option B describes acid-base neutralization, but while

NaOH

is indeed a base,

FeCl_3

isn't functioning as an acid here—it's simply providing

Fe^{3+}

ions for precipitation. Option D proposes single displacement with sodium replacing iron, but this would require sodium metal, not

NaOH

, and wouldn't occur anyway since iron is more reactive than sodium. Remember: when predicting reactions between ionic compounds, always consider solubility rules first. Metal hydroxides (except Group 1 and some Group 2) are typically insoluble, making precipitation reactions common when mixing metal cations with hydroxide ions.

Question 11

In the half-reaction $MnO_4^- \rightarrow Mn^{2+}$ in acidic solution, how many electrons are involved and on which side of the equation do they appear?

3 electrons on the left side
3 electrons on the right side
5 electrons on the left side (correct answer)
5 electrons on the right side
7 electrons on the right side

Explanation: When you encounter redox half-reactions, you need to balance both mass and charge by determining how many electrons are transferred and where they belong in the equation. To find the electrons in

MnO_4^- \rightarrow Mn^{2+}

, start by identifying the oxidation states. In

MnO_4^-

, manganese has an oxidation state of +7 (since oxygen is -2, and 4(-2) + 7 = -1 for the overall charge). In

Mn^{2+}

, manganese clearly has an oxidation state of +2. The change from +7 to +2 means manganese gains 5 electrons, so this is a reduction half-reaction. Since electrons are gained during reduction, they must appear as reactants on the left side of the equation:

MnO_4^- + 5e^- \rightarrow Mn^{2+}

. This confirms that 5 electrons appear on the left side. Looking at the wrong answers: A is incorrect because while it correctly places electrons on the left (reduction side), it miscounts the electron change as 3 instead of 5. B makes the same counting error and incorrectly places electrons on the right side, which would indicate oxidation rather than reduction. D correctly identifies 5 electrons but wrongly places them on the right side, again confusing this reduction with an oxidation. Remember this pattern: in reduction half-reactions, electrons always appear on the reactant side (left), while in oxidation half-reactions, they appear on the product side (right). Calculate the oxidation state change carefully to determine the exact number of electrons transferred.

Question 12

A student mixes 100.0 mL of 0.200 M $HCl$ with 50.0 mL of 0.400 M $NaOH$ . What is the concentration of the excess reagent in the final solution?

0.00 M (no excess) (correct answer)
0.033 M $HCl$
0.067 M $NaOH$
0.100 M $HCl$
0.133 M $NaOH$

Explanation: When you encounter acid-base neutralization problems, you need to determine whether the reactants are in stoichiometric proportions or if one is in excess. The balanced equation is

HCl + NaOH \rightarrow NaCl + H_2O

, showing a 1:1 molar ratio. First, calculate the moles of each reactant. For

HCl

: 0.1000 L × 0.200 M = 0.0200 mol. For

NaOH

: 0.0500 L × 0.400 M = 0.0200 mol. Since both reactants have exactly 0.0200 moles and react in a 1:1 ratio, they completely neutralize each other with no excess reagent remaining. The final solution contains only the salt

NaCl

and water, making the concentration of excess reagent 0.00 M, which is answer A. Answer B (0.033 M

HCl

) would be incorrect because it assumes

HCl

is in excess, but the calculations show equal molar amounts. Answer C (0.067 M

NaOH

) similarly assumes

NaOH

is in excess, which isn't the case. Answer D (0.100 M

HCl

) represents a significant miscalculation, perhaps confusing initial concentration with final concentration or forgetting about the neutralization reaction entirely. Study tip: Always start neutralization problems by calculating moles of each reactant, then compare to the stoichiometric ratio from the balanced equation. Only when moles are unequal will you have an excess reagent to calculate. Remember that "excess" means what's left over after complete reaction, not what you started with.

Question 13

In a titration, 25.0 mL of $H_2SO_4$ solution requires 40.0 mL of 0.150 M $NaOH$ for complete neutralization. What is the molarity of the sulfuric acid solution?

0.075 M
0.120 M (correct answer)
0.150 M
0.240 M
0.300 M

Explanation: When you encounter acid-base titration problems, you need to understand stoichiometry and the balanced chemical equation. The key insight here is that sulfuric acid is diprotic, meaning each molecule can donate two protons. First, write the balanced equation:

H_2SO_4 + 2NaOH → Na_2SO_4 + 2H_2O

. Notice the 1:2 molar ratio between sulfuric acid and sodium hydroxide. Calculate the moles of NaOH used: 0.0400 L × 0.150 mol/L = 0.00600 mol NaOH. Using the stoichiometric ratio, determine moles of

H_2SO_4

: Since 1 mole

H_2SO_4

neutralizes 2 moles NaOH, you have 0.00600 mol NaOH ÷ 2 = 0.00300 mol

H_2SO_4

. Finally, calculate molarity: 0.00300 mol ÷ 0.0250 L = 0.120 M. Choice A (0.075 M) represents a calculation error where someone might have incorrectly used the volume ratio (25.0/40.0) multiplied by the NaOH concentration, ignoring stoichiometry. Choice C (0.150 M) assumes a 1:1 ratio, forgetting that

H_2SO_4

is diprotic. Choice D (0.240 M) doubles the correct answer, perhaps from incorrectly multiplying by the stoichiometric coefficient instead of dividing. Remember: Always write the balanced equation first in titration problems. Polyprotic acids like

H_2SO_4

and

H_3PO_4

are common test topics, so practice identifying how many protons each acid can donate to avoid stoichiometric mistakes.

Question 14

Which statement best explains why the reaction $2Al(s) + 3CuSO_4(aq) \rightarrow Al_2(SO_4)_3(aq) + 3Cu(s)$ is classified as a single displacement reaction?

One compound breaks down into simpler substances
Two compounds exchange ions with each other completely
A more active metal displaces a less active metal from its compound (correct answer)
An acid and a base react to form a salt and water
Two elements combine to form a single compound

Explanation: When you encounter chemical reactions, identifying the reaction type helps predict products and understand the underlying chemistry. This question tests your ability to recognize single displacement reactions and understand the activity series of metals. In this reaction, aluminum metal is replacing copper in copper sulfate. The key insight is that aluminum is more reactive (more active) than copper, so it can displace copper from its compound. You can see this by noting that solid aluminum becomes part of a compound (

Al_2(SO_4)_3

), while copper goes from being in a compound to existing as a free metal. This is the hallmark of single displacement: one free element displaces another element from its compound. Looking at the wrong answers: (A) describes decomposition reactions, where one compound breaks into multiple simpler substances - but here we have a metal and a compound as reactants, not just one compound breaking down. (B) describes double displacement reactions, where two compounds swap ions completely (like

AB + CD \rightarrow AD + CB

) - but this reaction involves a free metal, not two compounds. (D) describes acid-base neutralization reactions that produce salt and water - this reaction involves neither acids nor bases, and no water is formed. The correct answer is (C) because aluminum, being higher on the activity series than copper, displaces the less active copper from its sulfate compound. Study tip: For single displacement reactions, remember the pattern: Active metal + Metal compound → Less active metal + New compound. Always check if the free element is more reactive than the one in the compound.

Question 15

What is the oxidation state of sulfur in the compound $Na_2S_2O_3$ (sodium thiosulfate)?

+1
+2 (correct answer)
+3
+4
+6

Explanation: When you encounter oxidation state problems, you need to use the fact that the sum of all oxidation states in a neutral compound equals zero, and apply known oxidation states for common elements. In

Na_2S_2O_3

, start with what you know: sodium (Na) always has an oxidation state of +1, and oxygen typically has an oxidation state of -2. Since there are 2 sodium atoms contributing +2 total and 3 oxygen atoms contributing -6 total, you have: 2(+1) + 2(S) + 3(-2) = 0. Solving this equation: +2 + 2S - 6 = 0, which gives 2S = +4, so S = +2. However, thiosulfate has a special structure where the two sulfur atoms are not equivalent - one sulfur is bonded to oxygen atoms (oxidation state +6) and the other is bonded to the first sulfur (oxidation state -2). The average oxidation state is (+6 + (-2))/2 = +2. Looking at the wrong answers: A) +1 would result from incorrectly assuming sulfur behaves like a typical metal. C) +3 might come from dividing the total positive charge needed (+4) by an incorrect assumption about sulfur distribution. D) +4 is the trap of assigning this total positive charge to just one sulfur atom instead of recognizing it's distributed between two sulfur atoms. Remember that in polyatomic ions like thiosulfate, atoms of the same element can have different oxidation states due to different bonding environments. Always consider the average when asked for "the" oxidation state of an element appearing multiple times.

Question 16

In the reaction $2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$ , what mass of $KCl$ is produced when 4.90 g of $KClO_3$ decomposes completely?

2.98 g (correct answer)
3.65 g
4.90 g
5.95 g
7.30 g

Explanation: This is a stoichiometry problem that requires you to convert from grams of reactant to grams of product using molar ratios. When you see a balanced chemical equation with mass calculations, think mole-to-mole conversions as your bridge between substances. Start by finding the molar masses:

KClO_3

= 39.1 + 35.5 + 3(16.0) = 122.6 g/mol, and

KCl

= 39.1 + 35.5 = 74.6 g/mol. Convert the given mass to moles: 4.90 g

KClO_3

÷ 122.6 g/mol = 0.0400 mol

KClO_3

. From the balanced equation, the molar ratio is 2:2 (or 1:1) between

KClO_3

and

KCl

, so 0.0400 mol

KClO_3

produces 0.0400 mol

KCl

. Finally, convert to grams: 0.0400 mol

KCl

× 74.6 g/mol = 2.98 g

KCl

. Answer A (2.98 g) is correct. Answer B (3.65 g) likely results from incorrectly using the 2:3 ratio between

KClO_3

and

O_2

instead of the 2:2 ratio with

KCl

. Answer C (4.90 g) assumes no mass change during the reaction, ignoring that oxygen gas is released. Answer D (5.95 g) suggests using an incorrect molar mass or ratio calculation. Remember the stoichiometry roadmap: grams → moles → mole ratio → moles → grams. Always check that your answer makes chemical sense—the product mass should be less than the reactant mass when gas is evolved.

Question 17

When 2.40 g of carbon reacts completely with oxygen to form carbon monoxide according to $2C(s) + O_2(g) \rightarrow 2CO(g)$ , what mass of oxygen is consumed?

1.60 g
3.20 g (correct answer)
4.80 g
6.40 g
9.60 g

Explanation: This is a stoichiometry problem that requires you to use molar relationships from a balanced chemical equation to find how much of one substance reacts with a given amount of another. Start by identifying what you know: 2.40 g of carbon reacts completely, and you need to find the mass of oxygen consumed. The balanced equation

2C(s) + O_2(g) \rightarrow 2CO(g)

shows that 2 moles of carbon react with 1 mole of oxygen gas. First, convert the given mass of carbon to moles:

2.40 \text{ g C} \times \frac{1 \text{ mol C}}{12.01 \text{ g C}} = 0.200 \text{ mol C}

Next, use the molar ratio from the balanced equation. Since 2 moles of C react with 1 mole of O₂, you have:

0.200 \text{ mol C} \times \frac{1 \text{ mol O}_2}{2 \text{ mol C}} = 0.100 \text{ mol O}_2

Finally, convert moles of oxygen to grams:

0.100 \text{ mol O}_2 \times \frac{32.00 \text{ g O}_2}{1 \text{ mol O}_2} = 3.20 \text{ g O}_2

This confirms answer B) 3.20 g is correct. Answer A) 1.60 g would result from incorrectly using 16.00 g/mol (atomic mass of oxygen) instead of 32.00 g/mol for O₂. Answer C) 4.80 g suggests using a 1:1 molar ratio instead of the correct 2:1 ratio. Answer D) 6.40 g combines both errors—wrong molar ratio and wrong molar mass. Remember: always check that your balanced equation coefficients match your molar ratios, and use molecular masses for compounds, not atomic masses for individual elements.

Question 18

A 0.500 g sample of an unknown metal reacts with excess hydrochloric acid to produce 0.560 L of hydrogen gas at STP. If the metal has a +2 oxidation state in the compound formed, what is the molar mass of the metal?

20.0 g/mol (correct answer)
24.3 g/mol
40.0 g/mol
48.6 g/mol
65.4 g/mol

Explanation: When you encounter gas evolution reactions with metals, you're dealing with stoichiometry that connects moles of gas produced to moles of metal consumed. The key insight is using the ideal gas law and balanced chemical equations. Since the metal has a +2 oxidation state, the reaction is:

\text{M} + 2\text{HCl} \rightarrow \text{MCl}_2 + \text{H}_2

. This shows a 1:1 mole ratio between metal and hydrogen gas produced. First, find moles of H₂ gas. At STP, one mole of any gas occupies 22.4 L, so:

\text{moles of H}_2 = \frac{0.560 \text{ L}}{22.4 \text{ L/mol}} = 0.0250 \text{ mol}

Since the stoichiometry is 1:1, you also have 0.0250 mol of metal. The molar mass is:

\text{Molar mass} = \frac{0.500 \text{ g}}{0.0250 \text{ mol}} = 20.0 \text{ g/mol}

Looking at the wrong answers: B) 24.3 g/mol results from incorrectly using 18.4 L/mol instead of 22.4 L/mol at STP. C) 40.0 g/mol comes from assuming a 1:2 ratio (thinking 2 moles of metal produce 1 mole of H₂, which reverses the actual stoichiometry). D) 48.6 g/mol combines both errors—wrong gas volume and wrong stoichiometry. The correct answer is A) 20.0 g/mol. Study tip: Always write the balanced equation first to establish the mole ratio, and remember that at STP, 1 mole of gas = 22.4 L. These gas stoichiometry problems follow a predictable pattern: gas volume → moles of gas → moles of reactant → molar mass.

Question 19

A solution is prepared by dissolving 15.0 g of $NaCl$ in enough water to make 250.0 mL of solution. What is the molarity of this solution?

0.600 M
1.03 M (correct answer)
2.56 M
4.11 M
6.00 M

Explanation: This question tests your ability to calculate molarity, one of the most fundamental concentration units in chemistry. When you see a problem asking for molarity with given mass and volume, immediately think: molarity equals moles of solute divided by liters of solution. To find the molarity, you need to convert the 15.0 g of NaCl to moles, then divide by the solution volume in liters. First, calculate moles of NaCl using its molar mass (58.44 g/mol):

\frac{15.0 \text{ g}}{58.44 \text{ g/mol}} = 0.257 \text{ mol}

. Next, convert the volume to liters: 250.0 mL = 0.2500 L. Finally, calculate molarity:

M = \frac{0.257 \text{ mol}}{0.2500 \text{ L}} = 1.03 \text{ M}

, confirming answer B. Choice A (0.600 M) results from incorrectly using the mass of NaCl (15.0) divided by the volume in mL (250), without any unit conversions. Choice C (2.56 M) comes from dividing grams by liters without converting to moles first (15.0 g ÷ 0.250 L). Choice D (4.11 M) represents dividing the mass by the molar mass incorrectly, likely multiplying instead of dividing somewhere in the calculation. Remember the molarity formula: M = mol/L. Always convert mass to moles using molar mass, and volume to liters. Write out each step clearly to avoid unit conversion errors, which are the most common mistakes in molarity problems.

Question 20

When balancing the equation $Al + O_2 \rightarrow Al_2O_3$ , what is the sum of all coefficients in the balanced equation?

6
7
9 (correct answer)
12
15

Explanation: When you encounter chemical equation balancing problems, you're applying the law of conservation of mass—atoms can't be created or destroyed, so you need equal numbers of each type of atom on both sides of the equation. Start by counting atoms in the unbalanced equation:

Al + O_2 \rightarrow Al_2O_3

. The left side has 1 aluminum and 2 oxygen atoms, while the right side has 2 aluminum and 3 oxygen atoms. To balance aluminum, you need 2 Al atoms on the left, giving you:

2Al + O_2 \rightarrow Al_2O_3

. Now for oxygen: the right side has 3 oxygen atoms, but

O_2

molecules come in pairs. To get 3 oxygen atoms on the left using

O_2

molecules, you need to find the least common multiple of 2 and 3, which is 6. This means you need

3O_2

(giving 6 oxygen atoms) on the left and

2Al_2O_3

(also giving 6 oxygen atoms) on the right. The balanced equation becomes:

4Al + 3O_2 \rightarrow 2Al_2O_3

. Adding all coefficients: 4 + 3 + 2 = 9. Choice A (6) likely comes from forgetting to double the

Al_2O_3

coefficient. Choice B (7) might result from incorrectly balancing one element. Choice D (12) could come from using unnecessarily large coefficients that maintain balance but aren't in lowest terms. Always verify your balanced equation by counting atoms of each element on both sides, and remember to use the smallest whole-number coefficients possible.